Respuesta :

Area of the given rhombus = 498.83 ft²

Solution:

Given figure is a rhombus ABCD.

BE = 12 ft and ∠BAE = 30°

Property of rhombus:

Diagonals bisect each other at right angles.

In ΔAEB, ∠BAE = 30°, ∠AEB = 90° and BE = 12 ft

[tex]$\sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}}[/tex]

[tex]$\sin30^\circ=\frac{\text{BE}}{\text{AB}}[/tex]

[tex]$\frac{1}{2} =\frac{\text{12}}{\text{AB}}[/tex]

Do cross multiplication, we get

AB = 24 ft

Using Pythagoras theorem,

[tex]\text {Adjacent}^{2}+\text{Opposite}^{2}=\text{ Hypotenuse }^{2}[/tex]

[tex]AE^2+BE^2=AB^2[/tex]

[tex]AE^2+12^2=24^2[/tex]

[tex]AE^2+144=576[/tex]

[tex]AE^2=432[/tex]

Taking square root on both sides, we get

[tex]AE=12\sqrt{3}[/tex] ft

AC = [tex]2\times 12\sqrt{3}=24\sqrt {3}[/tex]

Area of the rhombus = [tex]\frac{1}{2}\times d_1 \times d_2[/tex]

                                   [tex]$=\frac{1}{2}\times 24 \times 24\sqrt3[/tex]

                                   [tex]=288\sqrt{3}[/tex]

                                   = 498.83 ft²

Area of the given rhombus = 498.83 ft²

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